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QUESTION: 1

The masses of the three wires of copper are inthe ratio of 1 : 3 : 5 and their lengths are in theratio of 5 : 3 : 1. The ratio of their electricalresistance is [1988]

Solution:

QUESTION: 2

Two batteries of emf 4 V and 8V with internal resistance 1 Ω and 2 Ω are connected in a circuit with a resistance of 9 Ω as shown in figure. The current and potential difference between the points P and Q are [1988]

Solution:

QUESTION: 3

n equal resistors are first connected in seriesand then connected in parallel. What is the ratioof the maximum to the minimum resistance ?[1989]

Solution:

In series, R_{s}= nR

QUESTION: 4

Forty electric bulbs are connected in series across a 220 V supply. After one bulb is fused the remaining 39 are connected again in series across the same supply. The illumination will be [1989]

Solution:

Since, the voltage is same for the two

combinations, therefore

H ∝ 1/R.

Hence, the

combination of 39 bulbs will glow more.

QUESTION: 5

A current of 2 A, passing through a conductorproduces 80 J of heat in 10 seconds. Theresistance of the conductor in ohm is [1989]

Solution:

QUESTION: 6

Two identical batteries each of e.m.f 2V andinternal resistance 1Ω are available to produceheat in an external resistance by passing a currentthrough it. The maximum power that can bedeveloped across R using these batteries is [1990]

Solution:

For maximum current, the two batteries

should be connected in series. The current

will be maximum when external resistance is

equal to the total internal resistance of cells

i.e. 2Ω. Hence power devloped across the

resistacne R will be

QUESTION: 7

You are given several identical resistances eachof value R = 10Ω and each capable of carrying amaximum current of one ampere. It is required tomake a suitable combination of these resistancesof 5Ω which can carry a current of 4 ampere. Theminimum number of resistances of the type Rthat will be required for this job is [1990]

Solution:

To carry a current of 4 ampere, we need four

paths, each carrying a current of one ampere.

Let r be the resistance of each path. These

are connected in parallel. Hence, their

equivalent resistance will be r/4. According

to the given problem r/4 = 5 or r = 20Ω

For this propose two resistances should be

connected. There are four such

combinations. Hence, the total number of

resistance = 4 × 2 = 8.

QUESTION: 8

In the network shown in the Fig, each resistance is 1Ω. The effective resistance between A and B is [1990]

Solution:

At A current is distributed and at B currents

are collected. Between A and B, the

distribution is symmetrical. It has been

shown in the figure. It appears that current

in AO and OB remains same. At O, current i4

returns back without any change. If we

detach O from AB there will not be any

change in distribution.

Now, CO & OD will be in series hence its

total resistance = 2Ω

It is in parallel with CD, so, equivalent

resistance =

This equivalent resistance is in series with

AC & DB, so, total resistance

Now 8/3 Ω is parallel to AB, that is, 2Ω so

total resistance

Between C & D, the equivalent resistance is

given by

Equivalent resistance along

∴ Effective resistance between A and B is

QUESTION: 9

Faraday’s laws are consequence of conservationof [1991]

Solution:

Faraday’s laws are based on the conversion

of electrical energy into mechanical energy;

which is in accordance with the law of

conservation of energy

QUESTION: 10

The velocity of charge carriers of current (about1 amp) in a metal under normal conditions is ofthe order of [1991]

Solution:
The velocity of charge carriers of current 1 ampere in ametal under normal conditions is of the order of a fraction ofmm/sec

QUESTION: 11

Direct current is passed through a coppersulphate solution using platinum electrodes. Theelements liberated at the electrodes are [1993]

Solution:

In the electrolysis of CuSO_{4}, oxygen is

liberated at anode and copper is deposited

at cathode.

QUESTION: 12

Three resistances each of 4 Ω are connected toform a triangle. The resistance between any twoterminals is [1993]

Solution:

The two resistances are connected in series

and the resultant is connected in parallel with

the third resistance.

QUESTION: 13

Current through 3 Ω resistor is

0.8 amp., then potential drop through 4Ω resistor is [1993]

Solution:

Voltage across 3 Ω resistance = 3 × 0.8 = 2.4V

This voltage is the same across 6 Ω

resistance. Hence current through this

resistance

Total current in the circuit

= 0.8 + 0.4 = 1.2 amp

Voltage across 4 Ω resistance

= 4 × 1.2 = 4.8 volts

QUESTION: 14

A 4 μF capacitor is charged to 400 volts andthen its plates are joined through a resistance of1k Ω. The heat produced in the resistance is [1994]

Solution:

The energy stored in the capacitor

This energy will be converted into heat in

the resistor.

QUESTION: 15

In the circuit shown in Fig, the current in 4 Ω resistance is 1.2 A. What is the potential difference between B and C. [1994]

Solution:

The potential difference across 4Ω resistance

is given by

V = 4 × i_{1} = 4 × 1.2 = 4.8 volt

So, the potential across 8Ω resistance is also

4.8 volt.

Current in 2Ω resistance i = i_{1} + i_{2}

∴ i = 1.2 + 0.6 = 1.8 amp

Potential difference across 2Ω resistance

V_{BC} = 1.8 × 2 = 3.6volts

QUESTION: 16

A heating coil is labelled 100 W, 220 V. The coil iscut in half and the two pieces are joined in parallelto the same source. The energy now liberatedper second is [1995]

Solution:

Power of heating coil = 100 W and voltage

(V) = 220 volts. When the heating coil is cut

into two equal parts and these parts are

joined in parallel, then the resistance of the

coil is reduced to one-fourth of the previous

value. Therefore energy liberated per second

becomes 4 times i.e., 4 × 100 = 400 J.

QUESTION: 17

If a wire of resistance R is melted and recasted tohalf of its length, then the new resistance of thewire will be [1995]

Solution:

Initial resistance (R_{1}) = R; Initial length is

1and final length ( 2) = 0.5 . Volume of a

wire = . A. Since the volume of the wire

remains the same after recasting, therefore

We also know that resistance of a wire (R)

[Alt : When wires are drawn from same

volume but with different area of

cross-section, then

QUESTION: 18

Two wires of the same metal have same length,but their cross-sections are in the ratio 3:1. Theyare joined in series. The resistance of thickerwire is 10Ω. The total resistance of thecombination will be [1995]

Solution:

Length of each wire = ; Area of thick wire

(A_{1}) = 3A; Area of thin wire (A_{2}) = A and resistance of thick wire (R1) = 10 Ω

Resistance (if is constant)

The equivalent resistance of these two

resistors in series

= R_{1} + R, = 30 + 10 = 40Ω.

QUESTION: 19

If the resistance of a conductor is 5 Ω at 50ºCand 7Ω at 100ºC, then the mean temperature coefficient of resistance (of the material) is[1996]

Solution:

**Correct Answer :- a**

**Explanation : **We know that R_{T} = R_{0}(1+αT) where RT = resistance at temp T, R_{0} = resistance at temp T = 0^{o}C and α= temperature coefficient of resistance.

Here, 5 = R_{0}(1+α.50)...(1)

and 7 = R_{0}(1+α.100)...(2)

(2)/(1)

⇒ 7/5 = (1+100α)/(1+50α)

7+350α = 5+500α

150α=2

⇒α=2/150

=0.013/^{o}C

QUESTION: 20

What will be the equivalent resistance of circuitshown in figure between two points A and D [1996]

Solution:

Equivalent Circuit

Equivalent Resistance of circuit

QUESTION: 21

If a negligibly small current is passed through awire of length 15 m and of resistance 5Ω havinguniform cross-section of 6 × 10^{–7} m^{2} , thencoefficient of resistivity of material, is [1996]

Solution:

Given : Length of wire (l) = 15m

Area (A) = 6 × 10^{–7} m^{2}

Resistance (R) = 5Ω.

We know that resistance of the wire material

QUESTION: 22

Kirchoff’s first law, i.e. ∑i = 0 at a junction, dealswith the conservation of [1992, 1997]

Solution:

We know from the Kirchhoff 's first law that

the algebraic sum of the current meeting at

any junction in the circuit is zero (i.e. ∑i = 0)

or the total charge remains constant.

Therefore, Kirchhoff's first law at a junction

deals with the conservation of charge

QUESTION: 23

A (100 W, 200 V) bulb is connected to a 160Vpower supply. The power consumption would be [1997]

Solution:

Power = 100 W, Voltage = 200 V

Resistance of bulb =

When bulb is applied across 160 V,

Current in bulb = 160/400 A

Power consumption =

QUESTION: 24

There are three copper wires of length and cross sectional area . Inwhich case is the resistance minimum? [1997]

Solution:

QUESTION: 25

The current in the following circuit is [1997]

Solution:

Resistance of ACB, R' = 3Ω + 3Ω = 6Ω.

For net resistance between A and B; R' = 6Ω

and 3Ω are in parallel.

QUESTION: 26

Three equal resistors connected across a sourceof e.m.f. together dissipate 10 watt of power.What will be the power dissipated in watts if thesame resistors are connected in parallel acrossthe same source of e.m.f.? [1998]

Solution:

In series, Equivalent resistance = 3R

∴ Equivalent resistance R' = R/3

QUESTION: 27

If nearly 105 coulombs liberate 1 gm-equivalentof aluminium, then the amount of aluminium(equivalent weight 9), deposited throughelectrolysis in 20 minutes by a current of 50 amp.will be [1998]

Solution:

m = Zit, 9 = Z × 10^{5}, Z= 9 × 10–5

Again,

m = Zit = 9 × 10^{–5} × 50 × 20 × 60 = 5.4 gm

QUESTION: 28

A 5ºC rise in temperature is observed in aconductor by passing a current. When thecurrent is doubled the rise in temperature will beapproximately [1998]

Solution:

Since H ∝ I^{2} , doubling the current will

produce 4 times heat. Hence, the rise in

temperature will also be 4 times i.e., rise in

temperature = 4 × 5 = 20ºC.

QUESTION: 29

The resistance of a discharge tube is [1999]

Solution:

In discharge tube the current is due to flow

of positive ions and electrons. Moreover,

secondary emission of electrons is also

possible. So V-I curve is non-linear; hence

resistance is non-ohmic.

QUESTION: 30

The current (I) in the given circuit is [1999]

Solution:

In circuit, R_{B} and R_{C} are in series, so, Rs = 6 +

6 = 12Ω. This 12 Ω resistance is in parallel

with RA = 3 Ω,

So, equivalent resistance of circuit

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